# 5.3E: Nonhomgeneous Linear Equations (Exercises)

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- 18318

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In Exercises [exer:5.3.1} –[exer:5.3.6} find a particular solution by the method used in Example 5.3.2}. Then find the general solution and, where indicated, solve the initial value problem and graph the solution.

[exer:5.3.1] \(y''+5y'-6y=22+18x-18x^2\)

[exer:5.3.2] \(y''-4y'+5y=1+5x\)

[exer:5.3.3] \(y''+8y'+7y=-8-x+24x^2+7x^3\)

[exer:5.3.4] \(y''-4y'+4y=2+8x-4x^2\)

[exer:5.3.5] \(y''+2y'+10y=4+26x+6x^2+10x^3, \quad y(0)=2, \quad y'(0)=9\)

[exer:5.3.6] \(y''+6y'+10y=22+20x, \quad y(0)=2,\; y'(0)=-2\)

[exer:5.3.7] Show that the method used in Example

Example \(\PageIndex{1}\):

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.3.2} won’t yield a particular solution of

\[y''+y'=1+2x+x^2; \eqno{\rm (A)}\]

that is, (A) does’nt have a particular solution of the form \(y_p=A+Bx+Cx^2\), where \(A\), \(B\), and \(C\) are constants.[exer:5.3.8] \(x^2y''+7xy'+8y={6\over x}\) | [exer:5.3.9] \(x^2y''-7xy'+7y=13x^{1/2}\) |

[exer:5.3.10] \(x^2y''-xy'+y=2x^3\) | [exer:5.3.11] \(x^2y''+5xy'+4y={1\over x^3}\) |

[exer:5.3.12] \(x^2y''+xy'+y=10x^{1/3}\) | [exer:5.3.13] \(x^2y''-3xy'+13y=2x^4\) |

[exer:5.3.14] Show that the method suggested for finding a particular solution in Exercises [exer:5.3.8} -[exer:5.3.13} won’t yield a particular solution of

\[x^2y''+3xy'-3y={1\over x^3}; \eqno{\rm (A)}\]

that is, (A) doesn’t have a particular solution of the form \(y_p=A/x^3\).[exer:5.3.15] Prove: If \(a\), \(b\), \(c\), \(\alpha\), and \(M\) are constants and \(M\ne0\) then

\[ax^2y''+bxy'+cy=M x^\alpha\]

has a particular solution \(y_p=Ax^\alpha\) (\(A=\) constant) if and only if \(a\alpha(\alpha-1)+b\alpha+c\ne0\).[exer:5.3.16] \(y''+5y'-6y=6e^{3x}\) |
[exer:5.3.17] \(y''-4y'+5y=e^{2x}\) |

[exer:5.3.18] \(y''+8y'+7y=10e^{-2x}, \quad y(0)=-2,\; y'(0)=10\)

[exer:5.3.19] \(y''-4y'+4y=e^{x}, \quad y(0)=2,\quad y'(0)=0\)

[exer:5.3.20] \(y''+2y'+10y=e^{x/2}\) | [exer:5.3.21] \(y''+6y'+10y=e^{-3x}\) |

[exer:5.3.22] Show that the method suggested for finding a particular solution in Exercises [exer:5.3.16} -[exer:5.3.21} won’t yield a particular solution of

\[y''-7y'+12y=5e^{4x}; \eqno{\rm (A)}\]

that is, (A) doesn’t have a particular solution of the form \(y_p=Ae^{4x}\).[exer:5.3.23] Prove: If \(\alpha\) and \(M\) are constants and \(M\ne0\) then constant coefficient equation

\[ay''+by'+cy=M e^{\alpha x}\]

has a particular solution \(y_p=Ae^{\alpha x}\) (\(A=\) constant) if and only if \(e^{\alpha x}\) isn’t a solution of the complementary equation.[exer:5.3.24] \(y''-8y'+16y=23\cos x-7\sin x\)

[exer:5.3.25] \(y''+y'=-8\cos2x+6\sin2x\)

[exer:5.3.26] \(y''-2y'+3y=-6\cos3x+6\sin3x\)

[exer:5.3.27]\(y''+6y'+13y=18\cos x+6\sin x\)

[exer:5.3.28] \(y''+7y'+12y=-2\cos2x+36\sin2x, \quad y(0)=-3,\quad y'(0)=3\)

[exer:5.3.29] \(y''-6y'+9y=18\cos3x+18\sin3x, \quad y(0)=2,\quad y'(0)=2\)

[exer:5.3.30] Find the general solution of

\[y''+\omega_0^2y =M\cos\omega x+N\sin\omega x,\]

where \(M\) and \(N\) are constants and \(\omega\) and \(\omega_0\) are distinct positive numbers.[exer:5.3.31] Show that the method suggested for finding a particular solution in Exercises [exer:5.3.24} -[exer:5.3.29} won’t yield a particular solution of

\[y''+y=\cos x+\sin x; \eqno{\rm (A)}\]

that is, (A) does not have a particular solution of the form \(y_p=A\cos x+B\sin x\).[exer:5.3.32] Prove: If \(M\), \(N\) are constants (not both zero) and \(\omega>0\), the constant coefficient equation

\[ay''+by'+cy=M\cos\omega x+N\sin\omega x \eqno{\rm (A)}\]

has a particular solution that’s a linear combination of \(\cos\omega x\) and \(\sin\omega x\) if and only if the left side of (A) is not of the form \(a(y''+\omega^2y)\), so that \(\cos\omega x\) and \(\sin\omega x\) are solutions of the complementary equation.[exer:5.3.33] \(y''+5y'-6y=22+18x-18x^2+6e^{3x}\) (See Exercises [exer:5.3.1} and [exer:5.3.16}.)

[exer:5.3.34] \(y''-4y'+5y=1+5x+e^{2x}\) (See Exercises [exer:5.3.2} and [exer:5.3.17}.)

[exer:5.3.35] \(y''+8y'+7y=-8-x+24x^2+7x^3+10e^{-2x}\) (See Exercises [exer:5.3.3} and [exer:5.3.18}.)

[exer:5.3.36] \(y''-4y'+4y=2+8x-4x^2+e^{x}\) (See Exercises [exer:5.3.4} and [exer:5.3.19}.)

[exer:5.3.37] \(y''+2y'+10y=4+26x+6x^2+10x^3+e^{x/2}\) (See Exercises [exer:5.3.5} and [exer:5.3.20}.)

[exer:5.3.38] \(y''+6y'+10y=22+20x+e^{-3x}\) (See Exercises [exer:5.3.6} and [exer:5.3.21}.)

[exer:5.3.39] Prove: If \(y_{p_1}\) is a particular solution of

\[P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)\]

on \((a,b)\) and \(y_{p_2}\) is a particular solution of\[P_0(x)y''+P_1(x)y'+P_2(x)y=F_2(x)\]

on \((a,b)\), then \(y_p=y_{p_1}+y_{p_2}\) is a solution of\[P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)+F_2(x)\]

on \((a,b)\).[exer:5.3.40] Suppose \(p\), \(q\), and \(f\) are continuous on \((a,b)\). Let \(y_1\), \(y_2\), and \(y_p\) be twice differentiable on \((a,b)\), such that \(y=c_1y_1+c_2y_2+y_p\) is a solution of

\[y''+p(x)y'+q(x)y=f\]

on \((a,b)\) for every choice of the constants \(c_1,c_2\). Show that \(y_1\) and \(y_2\) are solutions of the complementary equation on \((a,b)\).